Analisis Sistem Tenaga 1

Buku Ajar Analisa Sistem Tenaga 1

V3’ { - (Y32.V2’ + Y34.V40 + Y36.V60)

{ – [(-30,769230 +

j46,153846)(0,0081 +

j 0,0082) + (-13,725490 + j21,568627) (1,0) + (-
55,081967 + j86,557377) (1,0)

{ – (-69,0565 +

j108,5043)

=
=
= 1,0104 – j0,8383 pu

KOREKSI

92

Buku Ajar Analisa Sistem Tenaga 1
| V3’| =

=
=
= 0,5640

V3’ = V3’

= 1,0104 – j0,8383

= 1,9503– j 1,6349 pu

4. BUS 4 = Load Bus (P-Q)

V4’ { - (Y43 . V3’ + Y45 . V50)

{ – [(-13,725490 +

j21,568627)( 1,9503– j 1,6349)+ (-11,029411 +

j18,382352)(1,0)

93

Buku Ajar Analisa Sistem Tenaga 1

={

=
=
= 1,4102+ j1,3050 pu

94

Buku Ajar Analisa Sistem Tenaga 1
KOREKSI :

V4’ [ - (Y43 . V30 + Y45. V50)

{–

{-

{ ×
}- -

{

{ }-

(38,2656 – j 54,0966)
95

Buku Ajar Analisa Sistem Tenaga 1

=

=

=

=

= 1,4072 + j 1,2983 pu

5. BUS 5 = Load Bus (P-Q)

V5’ { - (Y54 . V4’ + Y56 . V60)

{ – (-11,029411 +

j18,382352)( 1,4072 + j 1,2983)+ (-10,256410 +

j15,384615)(1,0)

={

96

Buku Ajar Analisa Sistem Tenaga 1

= –
=
= 1,2001+ j1,0985 pu

KOREKSI :

V5’ [ - (Y54 . V40 + Y56 . V60)

{

{-

{ ×
}-

97

Buku Ajar Analisa Sistem Tenaga 1

{

-

{ }-

(26,4377 – j 39,8920)

=

=

=

=
= 1,1986 + j 1,0932 pu

6. BUS 6 = Load Bus (P-Q)

V6’ { - (Y61 . V10 + Y62 . V2’ + Y63 . V3’ + Y65 . V5’)

98

Buku Ajar Analisa Sistem Tenaga 1

{ – (-16,393442 +

j19,672131)( 1,05 + j0) + (-9,873363 + j13,736853)( 0,0081 +
j 0,0082)+ (-55,081967 + j86,557377)
(1,4102+ j1,3050) + (-10,256410 + j15,384615) (1,1986 + j
1,0932)

={

+ ( 12,2933 + 16,8184)

=

=
= 1,1305+ j1,0621 pu

99

Buku Ajar Analisa Sistem Tenaga 1
KOREKSI :

V6’ [ - (Y61 . V10 + Y62 . V2’ + Y63 . V3’ + Y65 . V5’)

{–

+ ( 12,2933 + 16,8184) -

{

{×
-

{

–
{ }-

(107,8377 – j 150,1197)

=

100

Buku Ajar Analisa Sistem Tenaga 1

=

=

=
= 1,1305 + j 1,0612 pu

 Koreksi | dengan angka toleransi (0,0001)
| = | V2’ – V20|
= | (0,0081 + j 0,0082– (1,0 + j 0,0) |
= | - j 0,0082|
=
=
=
= 0,9918 < Toleransi ?

| = Tetap (Generator Bus)
| = | V4’ – V40|

= | (1,4072 + j 1,2983) – (1,0 + j 0,0) |
= | 0,4072 - j 1,2983|
=
=

101

Buku Ajar Analisa Sistem Tenaga 1
=
= 1,5596 < Toleransi ?

| = | V5’ – V50|
= | (1,1986 + j 1,0932) – (1,0 + j 0,0) |
= | 0,1986 - j 1,0932|
=
=
=
= 1,0749 < Toleransi ?

| = | V6’ – V60|
= | (1,1305 + j 1,0612) – (1,0 + j 0,0) |
= |0, 1305 - j 1,0612|
=
=
=
= 1,0531 < Toleransi ?

Koreksi dengan faktor percepatan (α = 1,6)
V1’ = 1,05 + j 0,0 (Nilai tetap karena Sluck Bus)

102

Buku Ajar Analisa Sistem Tenaga 1

V2’ = V20 + α V2

= (1,0 + j 0,0) + (1,6) ( - j 0,0082)

= (1,0 + j 0,0) + (-1,5870 - j0,0131)
= 2,5870 - j0,0131

V3’ = 1,9503– j 1,6349 (Nilai tetap karena Generator Bus)

V4’ = V40 + α V4
= (1,0 + j 0,0) + (1,6) (0,4072 - j 1,2983)
= (1,0 + j 0,0) + (0,6515 - j2,0772)
= 1,6515 - j2,0772

V5’ = V50 + α V5
= (1,0 + j 0,0) + (1,6) (0,1986 - j 1,0932)
= (1,0 + j 0,0) + (0,3177 - j 1,7491)
= 1,3177 - j 1,7491

V6’ = V60 + α V6
= (1,0 + j 0,0) + (1,6) (0, 1305 - j 1,0612)
= (1,0 + j 0,0) + (0,2088- j 1,6979)
= 1,2088 - j 1,6979

103

Buku Ajar Analisa Sistem Tenaga 1

BAB VIII. ANALISIS ALIRAN DAYA
NEWTON RAPHSON

Tujuan Instruksional Umum
Mahasiswa memahami persamaan umum aliran daya non linier,
memahami analisis aliran daya Newton Raphson.

Tujuan Instruksional
- Mahasiswa bisa menurunkan persamaan umum aliran daya aktif
dan reaktif di setiap bus dalam bentuk persamaan hybrid.
- Mahasiwa bisa melakukan analisis aliran daya dengan metode
Newton Raphson.

8.1. Prinsip Penyelesaian Metode Newton-Raphson

Gambar 8.1. Kurva persamaan non linier
X0 merupakan iterasi ke nol
X1 merupakan iterasi ke satu
X2 merupakan iterasi ke dua

104

Buku Ajar Analisa Sistem Tenaga 1

8.2. Persamaan/Fungsi Satu Variabel:

f (x)  0

Dengan menggunakan "deret taylor":

f (x)  f ( x0 )  1 df ( x0 ) ( x  x0 )  1 df 2 ( x0 ) ( x  x0 ) 2  ....
1! dx 2! dx
2

...  1 df n (x0 ) (x  x0 )n  0
n! dx n

Dengan pendekatan "Linier" :

f (x)  f (x0 )  df ( x0 ) ( x  x0 ) 0
dx

Sehingga diperoleh :

x1  x0  df f (x0 )
(x0 ) / dx

Atau dapat ditulis sebagai berikut:

x (1)  x(0)  df f (x(0) )
(x(0) ) / dx

x(0)  harga awal

x(1)  harga pada iterasi ke 1

Rumus untuk iterasi ke (k 1)

x(k 1)  x(k)  df f (x(k))
(x(k) ) / dx

105

Buku Ajar Analisa Sistem Tenaga 1
Contoh Penerapan Metode Newton-Raphson

F (x)  x3  64 Fungsi dengan satu variabel

F!(x)  3x2

x0  5 125  64
75
xn1  xn  xn x1   0,8133

xn  F(xn ) x1  5  0,8133  4,1867
F !(xn )
(4,1867)3  64
 x3  64 x2  3.(4,1867) 2  0,1785
3xn 2
x2  4,1867  0,1785  4,0082

Dan seterusnya untuk iterasi selanjutnya.

8.3. Penerapan Metode Newton Raphson Fungsi dengan 2
Variabel:

F1  x12  x22  5x1  0

F2  x12  x22 1,5x2  0

F1  2 x1  5; F1  2x2
x1 x2

F2  2x1; F2  2x2  1,5
x1 x2

x1(0)  3 dan x2(0)  3

F10  32  32  5.3  3

F20  32  32 1,5.3  4,5

F10  (2)(3)  5  1 F10 1(026)(3)  6
x1 x2
F20 F20

Buku Ajar Analisa Sistem Tenaga 1

Dihitung :

x1  1

x2  0,333

Jadi harga x11 dan x 2 :
2

x11  x10  x11  3 1  2

x12  x20  x12  3  0,333  2,667

Perhitunganiterasi 2 :

F11  22  2,6672  5.2  1,1129

F21  22  (2,667)2 1,5(2,667)  0,8876

F11  (2)(2)  5  1 F11  (2)(2,667)  5,334
x1 x2 107
F21  (2)(2)  4 F21  (2)(2,667)  1,5  3,834

Buku Ajar Analisa Sistem Tenaga 1

f1  f (x)  f (x1 )  0 1,1129  1,1129

f2  f (x)  f (x1)  0  0,8876  0,8876

1 53,3,83344.xx1222    1,1129 
   0,8876
 4 

Dihitung :

x12  0,5143

x22  0,3051

Jadi harga x12 dan x 2 :
2

x12  x11  x12  2  0,5143  1,4857

x22  x12  x22  2,667  0,3051  2,3619

108

Buku Ajar Analisa Sistem Tenaga 1

Iterasi X1 X2
0 3 3
1 2 2,667
. . .
. . .
7 1,0004 2,0003
8 1,00000189 2,00000149

Untuk contoh di atas dibutuhkan 8 kali iterasi
Dengan nilai akhir X1 =1,00 dan X2 =2,00

8.4. Elemen dari Jacobian Matrik:

 H ij  Pi Nij  Pi Vj 
  j Vj 
 
 Qi Qi 
 Jij   j Lij  Vj Vj 
 


. 
H 22 H 23 N23 2   P2 
 H33 N33   
 H 32 J33 L33 3    P3 
 V3 
 J32 Q3 

 V3 

H 22 H 23 N23 
Jacobian Matrik : H32 10N9 
H 33 33 
 J 32
J 33 L33 

Buku Ajar Analisa Sistem Tenaga 1

Elemen Jacobian Matrik masing-masing Bus

Type Bus Elemen Jacobian Matrik
Slack -
Generator Bus Hij
Load Bus Hij , Jij , Lij , Nij

8.5. Persamaan Aliran Daya Pada Masing-Masing Bus

N

Pi  Vi. Vj [Gij cos(i  j ) Bij sin( i  j )]
j 1
N

Qi  Vi. Vj [Gij sin( i  j )  Bij cos(i  j )]
j 1

Vi  Vi e ji
 Vi i

Yij  Gij  jBij

Latihan

110

Buku Ajar Analisa Sistem Tenaga 1
Penyelesaian Analisis Aliran Daya Metode Newton-Raphson

Gambar 8.2. Diagram Segaris Sistem Tenaga Listrik
Persamaan Non Linier:

111

Buku Ajar Analisa Sistem Tenaga 1

Persamaan Yang Digunakan Untuk Menghitung /V/ dan Ɵ Pada
Setiap Iterasi Adalah:

 P2 P2 P2  
 2 V3  
 3  V3 V3   
 P3 P3 P3 V3    2   P2 
   3   
  3  V3 .    P3 
 Q3 Q3 
2    V3 Q3 
3  V3  
 Q3   V3
 2


. 
H 22 H 23 N23 2   P2 
 H 33 N33   
 H 32 J33 L33 3    P3 
 V3 
 J32 Q3 

 V3 

Matrik Ybus:

112

Buku Ajar Analisa Sistem Tenaga 1

 4  j5 0  4  j5 
 4  j10  4  j10
YBUS   0  4  j10 8  j15 

 4  j5

P2  P2 {V2 V2 G22  V2 V3 G23 cos(2 3)  V2 V3 B23 sin(2 3)}

P3  P3 {V3 V1 G31 cos(3 1)  V3 V1 B31 sin(3 1)  V3 V2 G32 cos(3 2 )
 V3 V2 B32 sin(3 2 )  V3 V3 G33}

Q3  Q3 {V3 V1 G31 sin(3 1)  V3 V1 B31 cos(3 1)  V3 V2 G32 sin(3 2 )
 V3 V2 B32 cos(3 2 )  V3 V3 B33}

H 22  P2  V2 V3 G23{ sin( 2 3 )}.1 V2 V3 B23{cos(2 3 )}.1
2

N33  V3 P3  V3 {V1 G31 cos(3 1)  V2 G32 cos(3 2 )  2V3 G33
V3

 V1 B31 sin( 3 1)  V2 B32 sin( 3 2 )}

H 22  P2  V2 V3 G23{ sin( 2 3 )}.1 V2 V3 B23{cos(2  3 )}.1
2

 0  V2 V3 B23{cos(2 3 )}.1

 1,2491,010cos0  0.1

 11,249

H 23  P2   V2 V3 G23 sin( 2 3).(1)  V2 V3 B23 cos(2  3 ).(1)
3

 0  V2 V3 B23{cos(2 3 )}.(1)

 1,2491,010cos0  0.(1)

 11,249

 N23  V3 P2
V3  V3 0  V2 G23 cos(2 3 )  V2 B23.sin( 2 3 )

113

 V3 V2 G23 cos(2 3 )  V3 V2 B23.sin( 2 3 )

Buku Ajar Analisa Sistem Tenaga 1

H 33  P3   V3 V1 G31 sin( 3 1).(1)  V3 V2 G32 sin( 3 2 ).(1)
3

 V3 V1 B31 cos(3 1).(1)  V3 V2 B32 cos(3 2 ).(1)

 V3 V1 B31 cos(3 1).(1)  V3 V2 B32 cos(3 2 ).(1)

 (1,0)(1,0)(5) cos(0  0)(1)  (1,0)(1,1249)(10) cos(0  0)(1)

 16,249

 N33  V3P3
 V3  V3 V1 G31 cos(3 1)  V2 G32 cos(3 2 )  2V3 G33

 (1,0).(1,0).(4) cos(0  0)  (1,1249).(4).cos(0  0)  (2).(1,0).(8)

 7,501

J32  Q3  V3 V2 G32 cos(3 2 )(1)  V3 V2 B32 sin( 3 2 ).(1)
2
114

 V3 V2 G32 cos(3 2 ).(1)  0

Buku Ajar Analisa Sistem Tenaga 1

Sedangkan:

 ΔP2  P2  V2 V2 G22  V2 V3 G23cos(θ2  θ3 )  V2 V3 B23sin( θ2  θ3 )

 1,7  (1,1249)(1,1249)(4)  (1,1249)(1,0)(4)cos(0  0)
 (1,1249)(1,0)(10)sin (0  0) 


 1,137

P3  P3 {V3 V1 G31 cos(3 1)  V3 V2 G32 cos(3 2 )  V3 V3 G33

 V3 V1 B31 sin( 3 1)  V3 V2 B32 sin( 3 2 )}

 2  (1,0)(1,0)(4) cos(0  0)  (1,0)(1,1249)(4) cos(0  0) sin( 0  
(1,0)(1,0)(8)  (1,0)(1,0)(5)sin( 0  0)  (1,0)(1,1249)(10) 0)

 1,501

115

Buku Ajar Analisa Sistem Tenaga 1

Dalam Bentuk Matrik:


 4,499 
 11,249 11,249  2   1,137 
11,249 16,249 7,501  3   1,501
 4,499  8,499
13,751    V3   0,249 
 1,0 
 


 4,499 
 11,249 11,249  2   1,137 
11,249 16,249 7,501  3   1,501
 4,499  8,499
13,751    V3   0,249 
 1,0 
 

 11,249 11,249  4,499
Det  11,249 16,249 
 8,499 7,501   1009,7
 4,499
13,751 

116

Buku Ajar Analisa Sistem Tenaga 1

1,137 11,249  4,499

1,501 16,249 7,501

2  0,249  8,499 13,751  34,15 0,0338
1009,7 1009,7

Dengan cara yang sama :

3  0,056
V3  0,027

Contoh 2:

Data Saluran:

117

Buku Ajar Analisa Sistem Tenaga 1
Data Bus:

Base system: KVA Base: 20 MVA; KV Base: 20 KV
Hasil Perhitungan Matrik Ybus:

118

Buku Ajar Analisa Sistem Tenaga 1
Jacobian Matriks dari Sistem

H22 N22 H23 H24 N24 H25 N25 H26 N26 Δθ2 ΔP2
J22 L22 J23 J24 L24 J25 L25 J26 L26 ΔQ2
H32 N32 H33 H34 N34 H35 N35 H36 N36
H42 N42 H43 H44 N44 H45 N45 H46 N46 Δθ3 ΔP3
J42 L42 J43 J44 L44 J45 L45 J46 L46 Δθ4 ΔP4
H52 N52 H53 H54 N54 H55 N55 H56 N56
J52 L52 J53 J54 L54 J55 L55 J56 L56 = ΔQ4
H62 N62 H63 H64 N64 H65 N65 H66 N66 Δθ5 ΔP5
J62 L62 J63 J64 L64 J65 L65 J66 L66
ΔQ5
Δθ6 ΔP6

ΔQ6

Dimana :

119

Buku Ajar Analisa Sistem Tenaga 1

H ij  Pi ; N ij  Vj Pi
 j Vj

J ij  Qi ; Lij  Vj Qi
 j Vj

Persamaan Daya Aktif dan Reaktif Tiap Bus

Bus 2 i = 2 j = 2, 3, 6
P2 = |V2||V2| {G22 Cos (θ2 – θ2) + B22 Sin (θ2 – θ2)} + |V2||V3|
{ G23 Cos (θ2 – θ3) + B23 Sin (θ2 – θ3)} + |V2||V6| { G26
Cos (θ2 – θ6) + B26 Sin (θ2 – θ6)}
= V2V2 { G22 Cos (0) + B22 Sin (0)} + V2V3 { G23 Cos (θ2
– θ3) + B23 Sin (θ2 – θ3)} + V2V6 { G26 Cos (θ2 – θ6) +
B26 Sin (θ2 – θ6)}
= (V2V2 {G22 (1) + V2V2 . B22 (0)} + V2V3 { G23 Cos (θ2 –
θ3) + B23 Sin (θ2 – θ3)} + V2V6 {G26 Cos (θ2 – θ6) + B26
Sin (θ2 – θ6)}
= ((V2)2 . G22) + V2V3 { G23 Cos (θ2 – θ3) + B23 Sin (θ2 –
θ3)} + V2V6 { G26 Cos (θ2 – θ6) + B26 Sin (θ2 – θ6)}

Q2 = |V2||V2| { G22 Sin (θ2 – θ2) - B22Cos (θ2 – θ2)} + |V2||V3|
{ G23 Sin (θ2 – θ3) - B23 Cos (θ2 – θ3)} + |V2||V6| { G25 Sin
(θ2 – θ6) - B26 Cos (θ2 – θ6)}

120

Buku Ajar Analisa Sistem Tenaga 1

= V2V2 . G22 Sin (0) - B22 Cos (0) + V2V3 { G23 Sin (θ2 –
θ3) - B23 Cos (θ2 – θ3)} + V2V6 {G26 Sin (θ2 – θ6) - B26
Cos (θ2 – θ6)}

= (V2V2 . G22 (0) - V2V2 . B22 (1)} + V2V3 { G23 Sin (θ2 –
θ3) - B23 Cos (θ2 – θ3)} + V2V6 { G26 Sin (θ2 – θ6) - B26
Cos (θ2 – θ6)}

= ((-V2)2. B22) + V2V3 {G23 Sin (θ2 – θ3) - B23 Cos (θ2 –
θ3)} + V2V6 {G26 Sin (θ2 – θ6) - B26 Cos (θ2 – θ6)}

Bus 3 i = 3 j = 2, 3, 4, 6
P3 = |V3||V2| {G32 Cos (θ3 – θ2) + B32 Sin (θ3 – θ2)} + |V3||V3|
{G33 Cos (θ3 – θ3) + B33 Sin
(θ3 – θ3)} + |V3||V4| {G34 Cos (θ3 – θ4) + B34 Sin (θ3 –
θ4)} + |V3||V6| {G36 Cos (θ3 – θ6) + B36 Sin (θ3 – θ6)}
= V3 V2 {G32 Cos (θ3 – θ2) + B32 Sin (θ3 – θ2)} + V3V3
{G33 Cos (0) + B33 Sin (0)} +
V3V4 {G34 Cos (θ3 – θ4) + B34 Sin (θ3 – θ4)} + V3V6
{G36 Cos (θ3 – θ6) + B36 Sin (θ3 – θ6)}
= V3 V2 {G32 Cos (θ3 – θ2) + B32 Sin (θ3 – θ2)} + (V3V3 .
G33 (1) + V3V3. B33 (0) + V3V4 {G34 Cos (θ3 – θ4) + B34
Sin (θ3 – θ4)}

121

Buku Ajar Analisa Sistem Tenaga 1

= V3 V2 {G32 Cos (θ3 – θ2) + B32 Sin (θ3 – θ2)} + ((V3)2 . G33)
+ V3V4 {G34 Cos (θ3 – θ4) + B34 Sin (θ3 – θ4)} + V3V6 {G36
Cos (θ3 – θ6) + B36 Sin (θ3 – θ6)}

Q3 = |V3||V2| { G32 Sin (θ3 – θ2) - B32 Cos (θ3 – θ2)} + |V3||V3|
{ G33 Sin (θ3 – θ3) - B33
Cos (θ3 – θ3)} + |V3||V4| {G34 Sin (θ3 – θ4) - B34 Cos (θ3 –
θ4)} + |V3||V6| {G36 Sin (θ3 – θ6) - B36 Cos (θ3 – θ6)}

= V3 V2 { G32 Sin (θ3 – θ2) - B32 Cos (θ3 – θ2)} + V3V3 { G33
Sin (0) + B33 Cos (0)} + V3V4 { G34 Sin (θ3 – θ4) - B34 Cos
(θ3 – θ4)} + V3V6 { G36 Sin (θ3 – θ6) - B36 Cos (θ3 – θ6)}

= V3 V2 { G32 Sin (θ3 – θ2) - B32 Cos(θ3 – θ2)} + (V3V3 . G33
(1) + V3V3 . B33 (0)} + V3V4 {G34 Sin (θ3 – θ4) - B34 Cos
(θ3 – θ4)} + V3V6 {G36 Sin (θ3 – θ6) - B36 Cos (θ3 – θ6)}

= V3 V2 { G32 Sin (θ3 – θ2) - B32 Cos (θ3 – θ2)} - ((V3)2 . B33)
+ V3V4 {G34 Sin (θ3 – θ4) - B34 Cos(θ3 – θ4)} + V3V6 { G36
Sin (θ3 – θ6) - B36 Cos(θ3 – θ6)}

Bus 4 i = 4 j = 3, 4, 5
P4 = |V4||V3| { G43 Cos (θ4 – θ3) + B43 Sin (θ4 – θ3)} + |V4||V4|
{ G44 Cos (θ4 – θ4) + B44 Sin (θ4 – θ4)} + |V4||V5| { G45
Cos (θ4 – θ5) + B45 Sin (θ4 – θ5)}

122

Buku Ajar Analisa Sistem Tenaga 1

= V4V3 {G43 Cos (θ4 – θ3) + B43 Sin (θ4 – θ3)} + V4V4 {G44
Cos (0) + B44 Sin (0)} + V4V5 {G45 Cos (θ4 – θ5) + B45 Sin
(θ4 – θ5)}

= V4V3 {G43 Cos (θ4 – θ3) + B43 Sin (θ4 – θ3)} + (V4V4 . G44
(1) + V4V4. B44 (0)} + V4V5 {G45 Cos (θ4 – θ5) + B45 Sin
(θ4 – θ5)}

= V4V3 {G43 Cos (θ4 – θ3) + B43 Sin (θ4 – θ3)} + ((V4)2 . G44)
+ V4V5 {G45 Cos (θ4 – θ5) + B45 Sin (θ4 – θ5)}

Q4 = |V4||V3| {G43 Sin (θ4 – θ3) - B43 Cos (θ4 – θ3)} + |V4||V4|
{G44 Sin (θ4 – θ4) - B44Cos (θ4 – θ4)} + |V4||V5| {G45 Sin
(θ4 – θ5) - B45 Cos (θ4 – θ5)}

= V4V3 {G43 Sin (θ4 – θ3) - B43 Cos (θ4 – θ3)} + V4V4 {G44
Sin (0) - B44 Cos (0)} + V4V5 {G45 Sin (θ4 – θ5) - B45 Cos
(θ4 – θ5)}

= V4V3 {G43 Sin (θ4 – θ3) - B43 Cos (θ4 – θ3)} + (V4V4 . G44
(0) - V4V4. B44 (1) + V4V5 {G45 Sin (θ4 – θ5) - B45 Cos (θ4
– θ5)}

= V4V3 {G43 Sin (θ4 – θ3) - B43 Cos (θ4 – θ3)} - ((V4)2 . B44) +
V4V5 {G45 Sin (θ4 – θ5) - B45 Cos (θ4 – θ5)}

123

Buku Ajar Analisa Sistem Tenaga 1

Bus 5 i = 5 j = 4, 5, 6

P5 = |V5||V4| { G54 Cos (θ5 – θ4) + B54 Sin (θ5 – θ4)} + |V5||V5|
{ G55 Cos (θ5 – θ5) + B55 Sin (θ5 – θ5)} + |V5||V6| { G56 Cos
(θ5 – θ6) + B56 Sin (θ5 – θ6)}

= V5V4 {G54 Cos (θ5 – θ4) + B54 Sin (θ5 – θ4)} + V5V5 {G55
Cos (0) + B55 Sin (0)} + V5V6 {G56 Cos (θ5 – θ6) + B56 Sin
(θ5 – θ6)}

= V5V4 {G54 Cos (θ5 – θ4) + B54 Sin (θ5 – θ4)} + (V5V5 . G55
(1) + V5V5 . B55 (0) + V5V6 {G56 Cos (θ5 – θ6) + B56 Sin (θ5
– θ6)}

= V5V4 {G54 Cos (θ5 – θ4) + B54 Sin (θ5 – θ4)} + ((V5)2 . G55)
+ V5V6 {G56 Cos (θ5 – θ6) + B56 Sin (θ5 – θ6)}

Q5 = |V5||V4| { G54 Sin (θ5 – θ4) - B54 Cos (θ5 – θ4)} + |V5||V5|
{ G55 Sin (θ5 – θ5) - B55 Cos (θ5 – θ5)} + |V5||V6| { G56 Sin
(θ5 – θ6) - B56 Cos (θ5 – θ6)}

= V5V4 {G54 Sin (θ5 – θ4) - B54 Cos (θ5 – θ4)} + V5V5 {G55
Sin (0) - B55 Cos (0)} + V5V6 {G56 Sin (θ5 – θ6) - B56 Cos
(θ5 – θ6)}

= V5V4 {G54 Sin (θ5 – θ4) - B54 Cos (θ5 – θ4)} + (V5V5 . G55
(0) - V5V5 . B55 (1) + V5V6 {G56 Sin (θ5 – θ6) - B56 Cos (θ5 –
θ6)}

124

Buku Ajar Analisa Sistem Tenaga 1

= V5V4 {G54 Sin (θ5 – θ4) - B54 Cos (θ5 – θ4)} - ((V5)2. B55) +
V5V6 {G56 Sin (θ5 – θ6) - B56 Cos (θ5 – θ6)}

Bus 6 i = 6 j = 1, 2, 3, 5, 6

P6 = |V6||V1| {G61 Cos (θ6 – θ1) + B61 Sin (θ6 – θ1)} + |V6||V2|
{ G62 Cos (θ6 – θ2) + B62 Sin (θ6 – θ2)} + |V6||V3| { G63 Cos
(θ6 – θ3) + B63 Sin (θ6 – θ3)} + |V6||V5| { G65 Cos (θ6 – θ5) +
B65 Sin (θ6 – θ5)} + |V6||V6| { G66 Cos (θ6 – θ6) + B66 Sin
(θ6 – θ6)}

= V6V1 {G61 Cos (θ6 – θ1) + B61 Sin (θ6 – θ1)} + V6V2 {G62
Cos (θ6 – θ2) + B62 Sin (θ6 – θ2)} + V6V3 {G63 Cos (θ6 – θ3)
+ B63 Sin (θ6 – θ3)} + V6V5 {G65 Cos (θ6 – θ5) + B65 Sin
(θ6 – θ5)} + V6V6 {G66 Cos (0) + B66 Sin (0)}

= V6V1 {G61 Cos (θ6 – θ1) + B61 Sin (θ6 – θ1)} + V6V2 {G62
Cos (θ6 – θ2) + B62 Sin (θ6 – θ2)} + V6V3 {G63 Cos (θ6 – θ3)
+ B63 Sin (θ6 – θ3)} + V6V5 {G65 Cos (θ6 – θ5) + B65 Sin
(θ6 – θ5)} + (V6V6 . G66 (1) + V6V6 . B66 (0)}

= V6V1 {G61 Cos (θ6 – θ1) + B61 Sin (θ6 – θ1)} + V6V2 {G62
Cos (θ6 – θ2) + B62 Sin (θ6 – θ2)} + V6V3 {G63 Cos (θ6 – θ3)
+ B63 Sin (θ6 – θ3)} + V6V5 {G65 Cos (θ6 – θ5) + B65 Sin
(θ6 – θ5)} + ((V6)2 . G66)

125

Buku Ajar Analisa Sistem Tenaga 1

Q6 = |V6||V1| {G61 Sin (θ6 – θ1) – B61 Cos (θ6 – θ1)} + |V6||V2|
{G62 Sin (θ6 – θ2) – B62Cos (θ6 – θ2)} + |V6||V3| {G63 Sin
(θ6 – θ3) – B63 Cos (θ6 – θ3)} + |V6||V5| { G65 Sin (θ6 – θ5) –
B65 Cos (θ6 – θ5)} + |V6||V6| {G66 Sin (θ6 – θ6) - B66 Cos (θ6
– θ6)}

= V6V1 {G61 Sin (θ6 – θ1) – B61 Cos (θ6 – θ1)} +V6V2 {G62 Sin
(θ6 – θ2) – B62Cos (θ6 – θ2)} +V6V3 {G63 Sin (θ6 – θ3) – B63
Cos (θ6 – θ3)} + V6V5 {G65 Sin (θ6 – θ5) – B65 Cos (θ6–θ5)} +
V6V6 {G66 Sin (0) - B66 Cos (0)}

= V6V1 {G61 Sin (θ6 – θ1) – B61 Cos (θ6 – θ1)} + V6V2 {G62 Sin
(θ6 – θ2) – B62Cos (θ6 – θ2)} + V6V3 {G63 Sin (θ6 – θ3) – B63
Cos (θ6 – θ3)} + V6V5 {G65 Sin (θ6 – θ5) – B65 Cos (θ6 – θ5)}
+ (V6V6 . G66 (0) - V6V6 . B66 (1)}

= V6V1 {G61 Sin (θ6 – θ1) – B61 Cos (θ6 – θ1)} + V6V2 {G62 Sin
(θ6 – θ2) -B62Cos (θ6 – θ2)} + V6V3 {G63 Sin (θ6 – θ3) -B62Cos
(θ6 – θ3)} + V6V5 {G65 Sin (θ6 – θ5) – B65 Cos (θ6 – θ5)} -
((V6)2 . B66)

Proses Aliran Daya Iterasi 1

Elemen Jacobian Matriks

H 22  P2
 2

126

Buku Ajar Analisa Sistem Tenaga 1

= V2V3 (B23 Cos (θ2 – θ3) 1) + V2V6 (B26 Cos (θ2 – θ3) 1)
= 1,0 (1,1) (46,153846) 1 + (1,0)(1,0) (13,736853) 1
= 50,7692 + 13,7368
= 64,506

J 22  Q2
2

= V2V3 (G23 Cos (θ2 – θ3) 1) + V2V6 (G26 Cos (θ2 – θ6) 1)
= 1,0 (1,1) (-30,769230) 1 + (1.0) (1.0) (-9,873363) 1
= - 33,8461- 9,8733
= - 43,7194

L22  V2 Q2
 V2

L22 = |V2|

= – 2 |V2|B22 1– |V2||V3| B23 Cos (θ2 – θ3) 1 -|V2||V6| B26
Cos (θ2 – θ6) 1

= – 2 (1,0) (-59,890699) 1 – (1,0) (1,1) (46,153846) 1 -
(1,0) (1,0) (13,736853) 1

= 119,7813 - 50,7692 - 13,7368
= 135,2753

127

Buku Ajar Analisa Sistem Tenaga 1

N 22  V2 P2
 V2

= 2|V2| G22 1+ |V2||V3| G23 Cos (θ2 – θ3) 1 +|V2||V6| G26 Cos
(θ2 – θ6) 1}

= 2 (1,0) (40,642593) 1 + (1,0) (1,1) (-30,769230) 1 + (1,0)

(1,0) (-9,873363) 1}

= 81,2851 - 33,8461 - 9,8733

= 37,5657

H 23  Q2
3

= |V2||V3| {B23 Cos (θ2 – θ3) (-1)}

= (1,0) (1,1) (46,153846) (-1)

= 50,7692

J23 =

= |V2||V3| {G23 Cos (θ2 – θ3) (-1)}
= (1,0) (1,1) (- 30,769230) (-1)
= -33,8461
H24 = = 0

J24 = = 0

128

Buku Ajar Analisa Sistem Tenaga 1

L24 = |V4| =0

N24 = |V4| =0

H25 = = 0

J25 = = 0

L25 = |V5| =0

N25 = |V5| =0

H26 =

= |V2||V6| {B26 Cos (θ2 – θ3) (-1)}
= (1,0) (1,0) (13,736853) (-1)
= - 13,7368
J26 =

= |V2||V6| {G26 Cos (θ2 – θ3) (-1)}
= (1,0) (1,0) (-9,873363) (-1)
= 9,8733

129

Buku Ajar Analisa Sistem Tenaga 1
L26 = |V6|

= |V2| {-|V6| B26 Cos (θ2 – θ3) 1}
= (1,0) {-(1,0) (-13,7368) (1)}
= 13,7368
N26 = |V6|
= |V2| {|V6| G26 Cos (θ2 – θ6) 1}
= (1,0) {(1,0) (9,8733) 1}
= 9,8733
H32 =
= |V3||V2| {B23 Cos (θ3 – θ2) (-1)}
= (1,1) (1,0) (46,153846) (-1)
= - 50,7692
N32 =
= |V3||V2| {G23 Cos (θ3 – θ2)1}
= (1,1) (1,0) (– 30,769230) 1
= 33,8461

130

Buku Ajar Analisa Sistem Tenaga 1

H33 =

= V3V2 (B32 Cos (θ3 – θ2) 1) + V3V4 (B34 Cos (θ3 – θ4) 1) +
V3V6 (B36 Cos (θ3 – θ6) 1)

= (1,1) (1,0) (46,153846) 1 + (1,1) (1,0) (21,568627) 1+
(1,1) (1,0) (86,557377) 1

= 50,7692 + 21,5686 + 95,2131
= 167,5509

H34 =

= |V3||V4| {B34 Cos (θ3 – θ4) (-1)}
= (1,1) (1,0) (21,568627) (-1)
= - 23,7254

N34 = |V3|

= |V3| {|V4| G34 Cos (θ3 – θ4)}
= (1,1) (1,0) (–13,725490)}
= -15,0980

H35 = = 0

N35 = |V5| =0

131

Buku Ajar Analisa Sistem Tenaga 1

H36 =

= |V3||V6| {B36 Cos (θ3 – θ6) (-1)}
= (1,1) (1,0) (86,557377) (-1)
= 95,2131
N36 = |V6|

= |V6| {|V3| G36 Cos (θ3 – θ6)}
= (1,1) (1,0) (–55,081967)}
= 60, 5901
H42 = = 0

J42 = = 0

L42 = |V2| =0

N42 = |V2| =0

H43 =

= |V4||V3| {B43 Cos (θ4 – θ3) (-1)}
= (1,0) (1,1) (21,568627) (-1)
= -23,7254

132

Buku Ajar Analisa Sistem Tenaga 1

J43 =

= |V4||V3| {G43 Cos (θ4 – θ3) (-1)}
= (1,0) (1,1) (–13,725490) (-1)
= 15,0980
H44 =

= |V4||V3| (B43 Cos (θ4 – θ3) 1) + V4V5 (B45 Cos (θ4 – θ5) 1)
= (1,0) (1,1) (21,568627) (1) + (1,0) (1,0) (18,382352) (1)
= 23,7254 + 18,3823
= 42,1077
J44 =

= |V4||V3| {G43 Cos (θ4 – θ3) (1)} + |V4||V5| {G45 Cos (θ4 –
θ5) (1)}

= (1,0) (1,1) (-13,725490) (1) + (1,0) (1,0) (-11,029411) (1)
= -15,0980 -11,0294
= - 26,1274

L44 = |V4|

= |V4| {- |V3|B43 Cos (θ4 – θ3) 1 – 2 |V4|B44 1– |V4||V5| B45
Cos (θ4 – θ5) 1

= (1,0) {-(1,1) (21,568627) (1)} – 2 (1,0) (-39,950979) (1)
– (1,0)(1,0) (18,382352) (1)

133

Buku Ajar Analisa Sistem Tenaga 1

= -23,7254 + 79,9019 -18,3823
= 37,7942

N44 = |V4|

= |V4| {|V3| G43 Cos (θ4 – θ3)} (1) + 2 |V4| G44 (1) + |V4||V5|
{G45 Cos (θ4 – θ5) (1)

= (1,0) {(1,1) (-13,725490)1} + 2(1,0) (24,754901) (1) +
(1,0) (1,0) ( -11,029411) (1)

= -15,0980 + 49,5098 + (-11,0294)
= 23,3824

H45 =

= |V4||V5| {B45 Cos (θ4 – θ5) (-1)}
= (1,0)(1,0) (18,382352) (-1)
= -18,3823

J45 =

= |V4||V5| {G45 Cos (θ4 – θ5) (-1)}
= (1,0)(1,0) (– 11,029411) (-1)
= 11,0294

L45 = |V5|

= |V4| {-|V5|B45 Cos (θ4 – θ5) 1}
= (1,0){-(1,0) (18,382352) (1)}
= -18,3823

134

Buku Ajar Analisa Sistem Tenaga 1

N45 = |V5|

= |V4| {|V5|G45 Cos (θ4 – θ5) }
= (1,0){(1,0) (– 11,029411) }
= – 11,0294
H46 = = 0

J46 = = 0

L46 = |V6| =0

N46 = |V6| =0

H52 = = 0

J52 = = 0

L52 = |V2| =0

N52 = |V2| =0

H53 = = 0

J53 = = 0

H54 =

135

Buku Ajar Analisa Sistem Tenaga 1

= |V5||V4| { B54 Cos (θ5 – θ4) (-1)}
= (1,0)(1,0) (18,382352) (-1)
= -18,3823
J54 =

= |V5||V4| {G54 Cos (θ5 – θ4) (-1)}
= (1,0)(1,0) (– 11,029411) (-1)
= 11,0294
L54 = |V4|

= |V5| {-|V4| B54 Cos (θ5 – θ4) 1}
= (1,0){-(1,0) (18,382352) (1)}
= -18,3823
N54 = |V4|

= |V5| {|V4| G54 Cos (θ5 – θ4) }
= (1,0){(1,0) (– 11,029411) }
= -11,0294
H55 =

= |V5||V4| (B54 Cos (θ5 – θ4) 1) + V5V6 (B56 Cos (θ5 – θ6) 1)
= (1,0)(1,0) (18,382352) (1) + (1,0)(1,0) (15,384615) (1)
= 18,3823 + 15,3846
= 33,7669

136

Buku Ajar Analisa Sistem Tenaga 1

J55 =

= |V5||V4| { G54 Cos (θ5 – θ4) (1)} + |V5||V6| { G56 Cos (θ5 –
θ6) (1)}

= (1,0)(1,0) (-11,029411) (1) + (1,0)(1,0) (-10,256410) (1)
= (-10,2564) +(-10,2564)
= -21,2858
L55 = |V5|

= |V5| {- |V4|B54 Cos (θ5 – θ4) 1 – 2 |V5|B55 1– |V5||V6| B56
Cos (θ5 – θ6) 1

= (1,0){-(1,0) (18,382352) (1)} – 2 (1,0) (-33,766967) (1)
– (1,0)(1,0) (15,384615) (1)
= (-18,3823) + 67,5339 – 15,3846
= 33,767
N55 = |V5|

= |V5| {|V4| G54 Cos (θ5 – θ4)} (1) + 2 |V55| G55 (1) +
|V5||V6| { G56 Cos (θ5 – θ6) (1)

= (1,0){(1,0) (-11,029411)1} + 2(1,0)( 21,285821)(1) +
(1,0)(1,0)(-10,256410)(1)

= -11,0294 + 42,5716 + (-10,2564)
= 21,2858

137

Buku Ajar Analisa Sistem Tenaga 1

H56 =

= |V5||V6| {B56 Cos (θ5 – θ6) (-1)}
= (1,0)(1,0) (15,384615) (-1)
= -15,3846
J56 =

= |V5||V6| {G56 Cos (θ5 – θ6) (-1)}
= (1,0)(1,0) (– 10,256410) (-1)
= 10,2564
L56 = |V6|

= |V5| {-|V6| B56 Cos (θ5 – θ6) 1}
= (1,0){-(1,0) (15,384615) (1)}
= -15,3846
N56 = |V6|

= |V5| {|V6| G56 Cos (θ5 – θ6)}
= (1,0){(1,0) (– 10,256410) }
= -10,2564
H62 =

= |V6||V2| { B62 Cos (θ6 – θ2) (-1)}
= (1,0)(1,0) (13,736853) (-1)
= -13,7368

138

Buku Ajar Analisa Sistem Tenaga 1

J62 =

= |V6||V2| {G62 Cos (θ6 – θ2) (-1)}
= (1,0) (1,0) (– 9,873363) (-1)
= 9,8733
L62 = |V2|

= |V6| {-|V2| B62 Cos (θ6 – θ2) 1}
= (1,0) {-(1,0) (13,736853) (1)}
= -13,7368
N62 = |V2|

= |V6| {|V2| G62 Cos (θ6 – θ2)}
= (1,0) {(1,0) (– 9,873363)}
= -9,8733
H63 =

= |V6||V3| {B63 Cos (θ6 – θ2) (-1)}
= (1,0) (1,0) (86,557377) (-1)
= -86,5573
J63 =

= |V6||V3| G63 Cos (θ6 – θ2) (-1)}
= (1,0) (1,0) (–55,081967) (-1)
= 55,0819

139

Buku Ajar Analisa Sistem Tenaga 1

H64 = = 0

J64 = = 0

L64 = |V4| =0

N64 = |V4| =0

H65 =

= |V6||V5| {B65 Cos (θ6 – θ5) (-1)}
= (1,0) (1,0) (15,384615) (-1)
= -15,3846
J65 =

= |V6||V5| {G65 Cos (θ6 – θ5) (-1)}
= (1,0) (1,0) (– 10,256410) (-1)
= 10,2564
L65 = |V5|

= |V6| {-|V5| B65 Cos (θ6 – θ5) 1}
= (1,0) {-(1,0) (15,384615) (1)}
= -15,3846

140

Buku Ajar Analisa Sistem Tenaga 1

N65 = |V5|
= |V6| {|V5| G65 Cos (θ6 – θ5)}
= (1,0) {(1,0) (– 10,256410)}
= -10,2564

H66 =
= V6V1 (B61 Cos (θ6 – θ1) 1) + V6V2 (B62 Cos (θ6 – θ2) 1) +
V6V3 (B63 Cos (θ6 – θ3) 1) + V6V5 (B65 Cos (θ6 – θ5) 1)}
= 1,0 (1,05) (19,672131) 1 + 1,0 (1,0) (13,736853) 1 +
(1,0) (1,1) (86,557377) 1+ (1,0) (1,0) (15,384615) 1
= 20,6557 + 13,7368 + 95,2131 + 15,3846
= 144,9902

J66 =
= V6V1 (G61 Cos (θ6 – θ1) 1) + V6V2 (G62 Cos (θ6 – θ2) 1) +
V6V3 (B63 Cos (θ6 – θ3) 1) + V6V5(G65 Cos (θ6 – θ5)1}
= 1,0 (1,05) (-16,393442) 1 + 1,0 (1,0) (-9,873363) 1 +
(1,0) (1,1) (-55,081967) 1+ (1,0) (1,0) (-10,256410) 1
= (-17,2131) + (-9,8733) + (-60,5901) + (-10,2564)
= -97,9329

141